Question

如图，求角a+角b+角c+角d+角e

如图，求角a+角b+角c+角d+角e过程写完整

Answer 1

解:

设AE交CD于F,CD交AB于猜拿G,AB交孝罩DE于H

并设∠EFG=x,∠FGH=y,∠GHE=z

根据穗慎搭题意

b+c+y=180--(1)

a+e+z=180--(2)

d+e+x=180--(3)

x+y+z+e=360--(4)

(1)+(2)+(3),得

a+b+c+d+2e+x+y+z=540--(5)

(5)-(4),得

a+b+c+d+2e+x+y+z-(x+y+z+e)=a+b+c+d+e=540-360=180

故答案,180度

Answer 2

如图，由三角形外角和内角的关哗拦誉乱系，∠1=∠B+∠C，∠2=∠D+∠E。所以，∠A+∠B+∠C+∠D+∠E=∠A+(∠B+∠C)+(∠D+∠E)=∠A+∠1+∠乱虚胡2=180°

Answer 3

图液腊蠢闹陪局侍

追问

你好请问角dad是啥

追答

应该是<DAB

Answer 4

180°

Answer 5